Question

What are the points on the $x-$axis whose perpendicular distance from the line $\frac{x}{3}+\frac{y}{4}=1$is $4$units?

Answer 1

Step 1. Convert given equation of line in linear form.

Equation of line is,

$$\begin{gathered} \therefore \frac{x}{3}+\frac{y}{4}=1 \\ \Rightarrow \frac{4x+3y}{12}=1 \\ \Rightarrow 4x+3y=12 \\ \Rightarrow 4x+3y-12=0 \end{gathered}$$

Compare above equation with $Ax+By+Cy=0$.

We get, $A=4,B=3,C=-12$

Step 2. Find out the point on the $x-$axis.

The point on $x-$axis will be $(x,0)$.

The perpendicular distance of a point from a line is compuded as,

$$\begin{gathered} d=\frac{\left|Ax+By+C\right|}{\sqrt{A^2+B^2}} \\ \Rightarrow 4=\frac{\left|4x+3·0-12\right|}{\sqrt{16+9}}[\because d=4(\mathrm{Given}\left)\right] \\ \Rightarrow 4=\frac{4x-12}{\sqrt{25}} \\ \Rightarrow 4=\frac{\left|4x-12\right|}{5} \\ \Rightarrow \left|4x-12\right|=20 \\ \Rightarrow 4x-12=20\mathrm{or}4x-12=-20 \\ \Rightarrow x=\frac{32}{4}\mathrm{or}x=\frac{-20+12}{4} \\ \Rightarrow x=8\mathrm{or}x=-2 \end{gathered}$$

Hence, the required points are $(8,0)$ and $(-2,0)$.